WEBVTT
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This is College Physics
Answers with Shaun Dychko.
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A skier descends
a 70 meter slope
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which is inclined at an
angle of thirty degrees
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and we have to find out what their
final speed will be at the bottom,
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assuming there is no friction
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and assuming they have an initial
speed of zero for part A.
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Then we do the calculation again
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assuming an initial speed of 2.5
meters per second in part B.
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Then we compare the
total time it takes
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to get to the bottom of
the slope in each case.
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So we know that the total energy at
the end, kinetic plus potential,
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equals the total energy
at the beginning.
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So we have no potential
energy at the end
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because we'll assume this is our
reference level, y equals zero,
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so the final potential
energy is zero.
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So we have only kinetic energy
when the skier is at the bottom.
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So that's one half mass
times final speed squared,
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and in the initial case
we have no kinetic energy
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because we assume an initial
speed of zero for part A
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and the initial potential
energy will be mg times h,
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the vertical height
above the ground.
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So we can divide both sides by m
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and then multiply
both sides by two
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and also take the square
root of both sides.
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We end up with the final speed is the
square root of two g times the height.
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Now the height is the opposite
leg of this yellow triangle.
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So we go sine theta
multiplied by the hypotenuse
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which is d in order
to find the height.
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We'll substitute that in for h.
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We have vf equals square
root of twogd sine theta.
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So that's the square root of two
time 9.8 meters per second squared
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times 70 meters times sine 30
which is 26.2 meters per second.
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So that's the final speed
at the bottom of the slope.
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Then the next part
of this question is
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what time does it take to get
to the bottom of the slope.
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Well, the total displacement
along the slope
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is going to equal the average
velocity multiplied by the time.
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So we can solve this for t because we know
all these other things in the formula.
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So we'll say, let's multiply both
sides by two over v i plus vf
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and then switch
the sides around.
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We get that t is two
d over v i plus vf.
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So that's two times 70 meters
divided by zero initial speed,
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plus 26.2 meters per
second, final speed,
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giving us 5.35 seconds
to descend the slope.
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Now, in part B
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the only difference is that there
is an initial kinetic energy.
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So we have this one
half m visquared term
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which is the only difference compared
to that second line in part A.
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We will multiply both
sides by two over m
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and the two and the m cancel in the first
term leaving us with vi squared there.
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Then on the second
term the m's cancel,
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but we're left with the two behind.
So that's two gh there,
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and then we take the square root
of both sides to solve for vf.
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So vf equals the square
root of vi squared
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plus twog d sine theta
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where we substituted d sine
theta in place of the height h.
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Then we substitute in numbers.
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So it's the square root of 2.5 meters
per second initial speed squared,
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plus two times 9.8 times
70 times sine 30,
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giving us 26.3 meters per
second is the final speed
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which is nearly the same as the
final speed we had in part A, 26.2.
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But there is a bit of a --
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so this difference in
speeds is one tenth
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whereas the difference in times is
actually going to be more significant.
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It's still a small difference
but it's more significant.
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So we have the time is
going to be two d over vi
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plus vf for the same reason
that it was over here.
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We solved this
formula here for t
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and that's two times 70 divided by
2.5 meters per second initial speed,
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plus 26.3106 meters per
second final speed.
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That's 4.86 seconds is the total
time to get down the slope.
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So the difference
in times is small,
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5.35 minus 4.86 is only
about half a second.
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We expect that since the initial speed
is so much less than the final speed.
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But half a second could
make a big difference
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in placement be it first,
second, third, fourth
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in a highly competitive event.